Let $\beta \mathbb{N}$ be the Stone-Cech compactification of the natural numbers $\mathbb{N}$, and let $x, y \in \beta \mathbb{N} \backslash \mathbb{N}$ be two non-principal elements of this compactification (or equivalently, $x$ and $y$ are two non-principal ultrafilters). I am interested in ways to "model" the ultrafilter $y$ using the ultrafilter $x$. More precisely,

Q1. (Existence) Does there necessarily exist a continuous map $f: \beta \mathbb{N} \to \beta \mathbb{N}$ which maps $x$ to $y$, while mapping $\mathbb{N}$ to $\mathbb{N}$? To put it another way: does there exist a function $f: \mathbb{N} \to \mathbb{N}$ such that $\lim_{n \to x} f(n) = y$?

Q2. (Uniqueness) Suppose there are two continuous maps $f, g: \beta \mathbb{N} \to \beta\mathbb{N}$ with $f(x)=g(x)=y$, which map $\mathbb{N}$ to $\mathbb{N}$. Is it then true that $f$ and $g$ must then be equal on a neighbourhood of $x$?

I suspect the answer to both questions is either "no" or "undecidable in ZFC", though perhaps there exist "universal" ultrafilters $x$ for which the answers become yes. But I do not have enough intuition regarding the topology of $\beta \mathbb{N}$ (other than that it is somewhat pathological) to make this more precise. (The fact that $\beta\mathbb{N}$ is not first countable does seem to indicate that the answers should be negative, though.)

2 Answers
2

The answer to Q1 is no. This has been well studied in set theory; you're basically asking whether any two non-principal ultrafilters on $\mathbb{N}$ are comparable under the Rudin-Keisler ordering. Variations on your question have led to many, many interesting developments in set theory, but your question Q1 is easy to answer by a cardinality argument.

First note that every $f:\mathbb{N}\to\mathbb{N}$ has a unique extension to a continuous function $\bar{f}:\beta\mathbb{N}\to\beta\mathbb{N}$. Any $x \in \beta\mathbb{N}$ has at most $2^{\aleph_0}$ images through such $\bar{f}$, but there are $2^{2^{\aleph_0}}$ ultrafilters on $\mathbb{N}$, so there are very many $y \in \beta\mathbb{N}$ which are not images of $x$ through such $\bar{f}$.

The answer to Q2 is also no. Let $y$ be a nonprincipal ultrafilter on $\mathbb{N}$. The sets $A \times A\setminus\Delta$ where $A \in y$ and $\Delta = \{(n,n) : n \in \mathbb{N}\}$ form a filter base on $\mathbb{N}\times\mathbb{N}$. Let $x$ be an ultrafilter on $\mathbb{N}\times\mathbb{N}$ that contains all these sets. The left and right projections $\pi_1,\pi_2:\mathbb{N}\times\mathbb{N}\to\mathbb{N}$ both send $x$ to $y$, but they are not equal on any neighborhood of $x$.

However, the answer to Q2 is yes when $x$ is a selective ultrafilter. Recall that $x$ is selective if for every $h:\mathbb{N}\to\mathbb{N}$ there is a set $A \in x$ on which $h$ is either constant or one-to-one. Given $f,g:\mathbb{N}\to\mathbb{N}$ such that $\bar{f}(x) = \bar{g}(x)$ is nonprincipal, then we can find $A \in x$ on which $f$ and $g$ are both one-to-one. In that case, $f\circ g^{-1}$ must be well-defined on some $A' \in x$. Any extension of $f\circ g^{-1}$ to the complement of $A'$ must map $x$ to $x$, which means that $f \circ g^{-1}$ is the identity on some $A'' \in x$. Thus $f$ and $g$ are equal on the neighborhood of $x$ defined by $A''$.

I don't understand your cardinality argument - there are very many ultrafilters, yes, but there are only $2^{\alpha_0}$ choices that each ultrafilter can pick for $y$, as far as I can tell. What am I not seeing?
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Jacques CaretteApr 16 '10 at 20:38

As I understand François's argument: the cardinality of $\beta\mathbb{N}$ is bigger than the cardinality of the set of continuous functions $\beta\mathbb{N}\to\beta\mathbb{N}$ taking $\mathbb{N}\to\mathbb{N}$. So given $x$ you can find $y$ so that no such function takes $x$ to $y$.
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Tom ChurchApr 16 '10 at 20:50

The answer to Q1 is even more `no': Kunen showed that there are x and y such that x cannot be mapped to y and y cannot be mapped to x, see this review. This has been strengthened by Rudin and Shelah. It is still open, as far as I know, whether given x there is a y such that neither can be mapped to the other.

I maybe miss something, as I don't see how this question can be open: if you consider the oriented labeled graph with the Stone-Cech boundary as set of vertices, and oriented edge $(x,y)$ labeled by $f:I\to J$ whenever $f(x)=y$, and consider the underlying unoriented graph, then its valency is at most continuum, and therefore its components have cardinality at most continuum. So given $x$ you can find $y$ in another component.
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YCorOct 25 '12 at 18:57

The valency of the graph is $2^{\mathfrak{c}}$, rather than $\mathfrak{c}$; all points of the remainder have have $2^{\mathfrak{c}}$ preimages under non-trivial finite-to-one maps from $\mathbb{N}$ to itself (those where the fibers have arbitrarily large finite cardinality).
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KP HartOct 29 '12 at 20:35